Good afternoon, everybody, So, we are in the last class of coordination compounds where we are talking about the complexes and how we can explain their different physical behavior, as well as the property and in terms of valence bond theory, as well as the crystal field theory, we are trying to figure out the electronic configuration? So, we all know how we can have the corresponding electronic configuration and that the changed electronic configuration in say crystal field, where we see that a metal ion is considered as a positively charged point center. And the ligand is slightly bigger, negatively charged species. And this particular interaction when they are forming a coordinate bond. We consider them as interaction.

What we can find in the crystal of the salts like say, sodium chloride and how the different orbitals, particularly the d electron configuration. So, we are looking at the corresponding d electron configurations in the changed condition with regard to the corresponding geometries, which is very important how we can have a metal complex, whether we have a ML4, complex or ML6 complex.

What are the corresponding geometries? We should all know, and those geometries will also tell us that how many number of unpaired electrons will be there in that particular molecule or the coordination complex, because these number of unpaired electrons, which we should have some importance, related to their reactivity catalysis. And all these things, which we do not see very much in the corresponding organic chemistry but the coercion, chemistry and the inorganic chemistry of the transition metal ions, are largely dominated by the number of unpaired electrons. So those unpaired electrons have some important role to play and how we can determine, based on this crystal field theory, how we can determine the new n values. That means the number of n values which will be directly related to the corresponding magnetic moment values.

So, what we see now that we have five d levels or the d orbitals, and we know that in presence of say, six ligands their overall energy is elevated. And after that, from this value, which we consider as the e which is before the splitting. So, we require something which we consider crystal field. Speeding, so the splitting will be there and that splitting will give rise to a doublet state and a triplet state. And these two doublet states and triplet states are separated by say, a magnitude of x and the magnitude of y. So, if all of them that means the available capacity of all the five d orbitals. That means we can have ten electrons. So before splitting the energy will be 10 e and after spreading, we see that these are the two levels which can occupy four electrons, so it will be 4e+x. So, this will be 4e+x, 6e-y, so that basically give rise to 2 x is equivalent to 3y. So, what is the gap? Basically, we are looking at this gap from the level e. What is this gap x and what is that gap y, so this will give you some relationship, which is 2x=3y after splitting, so, on the left-hand side, we have the before speeding condition and, on the right hand, side after speeding and these two, we all know that this is the crystal field speeding, so this particular crystal field, speeding in octahedral geometry, which will be Δo, so that Δo will be this, and therefore we get the magnitude of x as and the magnitude of y. As and - and if this is also equal to sometime, we also write that is equivalent to 10 dqo, so will be equal to 6 dqo and will be equal to 4. These are the magnitude of these separations.

So, what we see that we know that this is the t2g level, and this is the eg level, so this t2g level is therefore stabilized by -4dqo this is the magnitude, the y, the magnitude of y and the e g level will therefore be destabilized by +6dqo, with respect to the unsplit very centers, with respect to this e level. So, with respect to this, how this splitting can take place we can find out now that the condition has changed, what we find is that we have a new electronic configuration. So, instead of saying that we have a dn electron configuration, that means the number of d electrons in the different levels. It can be three d, it can be four d or it can be five d levels, but now we just try to write in this fashion.

In octahedral geometry, what are the numbers in the t2g level, and what are the numbers in the e g level? So, we will designate over here the number of electrons in the t2g level. That means the corresponding level, which has a threefold degeneracy and, as I told you, that g is the corresponding term related to grad (a) which is equivalent to even - and this two is coming from a typical c two operation, because all the d orbitals why it Is even because all the d orbitals, what we see they are corresponding orbitals retain or they keep their sign of their wave function same with respect to the operation inversion. So that is the thing with respect to the inversion, which is at the middle and center. So that is also important that how the metal ion is there in the octahedral symmetry, so symmetry is octahedral, so all the orbitals, whatever orbitals we are talking about, they all follow the thing that they remain. The inverted form and the sign of those orbitals are maintained and they are of even category or grade category. So, this particular information will give rise to some other information to us, but we get that the crystal field, speeding, which can account for magnetic properties in terms of high spin and low spin complexes. So, these high spin and low spin complexes.

What we see here is related to the corresponding occupancy of the level t2g and eg, and now, if we just define how we can just go to that particular stage, that some of the ligands will categorize them as weak field, ligands and others are high spin. So weak field ligands are therefore giving rise to the corresponding high spin paramagnetic species. That means ligands are not changing the corresponding pairing of the levels with respect to the corresponding crystal field spitting. So, we see compared to that. We can have another category of ligands, which are strong field ligands, which will be responsible for low spin or rather diamagnetic systems. So, this low spin or di-aminetic systems are very important because sometime we find that the medallion electronic configuration is not immediately giving a low spin configuration of zero magnetic moment is typically diamante situation, but through oxidation, such as Co, , 2+ in whatever Coordination: geometry: whether it is a tetrahedral one or an octahedral, one is not diamagnetic, but once the octahedral species is oxidized for one electron giving rise to the corresponding trivalent cobalt center, we get a diamagnetic situation where the cobalt complexes would be diamagnetic. So that will see how it can be applied to a situation where we see that both of them are trivalent cobalt center.

In the left hand, side what we see that the fluoride ligand, so fluoride ligands, which is hexa- fluoro, covalent three species and, on the right hand, side, hexa-cyano cobalt, three species one both of them having charge on the ligands and the both of the complexes are Anionic, but the situation is such that, on the left hand, side our compounds are corresponding compounds are paramagnetic, because if you have, if you have six electrons to distribute for - and this is the t2g set - and the upper two are the eg set and since the Δ is small, so is the fluoride ligand. We can categorize this fluoride ligand as a weak field ligand. So, this fluoride ligand will be a weak field ligand in terms of the corresponding previous definition for the valence bond fixture.

What we have seen that having four unpaired electrons in the 3d6 configuration - that means all the 6 electrons will be paired up, and this can be defined from sp3d2 hybridizations. So, this sp3d2 type of hybridization - will give rise to high spin complexes, whereas for c o c n whole six, three minus or unpaired electron is zero, which is still 3d. Six electronic configuration and the hybridization is d. Two s v three and is low spin. So these two levels, what we are now bringing is not due to the valence bond picture, but is due to the crystal field picture, because in both these two cases, what we have seen that it is three d six and it is also three d: six and Antenna unless we divide these two into the corresponding t, two g set and the e g set, we cannot explain the different number of unpaired electrons for these two cases, so that will see here in the case of this high spin and the low spin complexes for the different Δ values so for the left, what we see here is the corresponding value for is less so.

The fluoride ligands are weak field ligands, so that give rise to a very small value of Δ, whereas the cyanide ligands give rise to the strong field for the metal ion and which, having some Δ value, which is pretty high. So simply just comparing these Δ values. The magnitude of these Δ values after crystal field, speeding, due to the presence of this ligand, we see that the cyanide c n minus will be a stronger ligand or a strong field ligand compared to fluoride. So, if we take these two examples, like that, what we have seen in case of the nickel, that water molecules, the ammonia molecules and ethylene diamine, simply we can talk in terms of the corresponding Δ values, and these are Δ o. That means the Δ for the octahedral symmetry, so we can have some leveling for these things and we just simply write now that instead of 3d, six electronic configurations for cobalt, three plus in an octahedral field.

If we have a configuration like this, how can we measure the corresponding stabilization due to crystal field, speeding, which we consider it as the crystal field, splitting energy. So, we will have a number of electrons in the t2g level and some other number of electrons in the eg level, and that will now give rise to the corresponding electronic configuration in terms of the crystal field picture. So, the net energy of the configuration related to the average energy of the orbital is what we have just now calculated.

Is that minus point four x plus point four six y this we are just calculating it out by two fifth of Δo and three fifth of Δo in respect to the Δ zero, so that will give rise to the Δo is ten dq and therefore what we see is a situation beyond d 3, because d, 1 d, 2 d, 3 3 electrons - will initially be fed into the t 2g level. So, there is no such competition for occupying the other level. But when we move to the default situation, the configuration is d4 so when the configuration is d4 in the weak field condition.

So, this is the statement that means in the weak field condition our Δ value is less than p. P is nothing but our pairing energy, whether this fourth electron, when we go beyond d3, whether the fourth electron will come to the t2g level or it will go to the eg level. So, there will be a choice so that choice will be dictated by the amount of splitting whether your Δ g, zero Δ o is applicable compared to p. If it is higher than p, this electron will come to the lower level, otherwise it will be in the easy level. So, the weak field situation will give rise to something where we have four electrons in these levels: three t2g level and one eg level. So, all of them will be unpaired, so we get some a situation which is a high spin situation, but for the strong field ligand, which we just now seen for the cyanide ligand the cyanide group, which is binding to the cobalt tree center, your Δ>p is greater than the pairing energy, so this Δo will be higher. So that is why the fourth electron will come to the t2g level, which is t2 g4. So simply this particular information can give rise to some information related to the electronic configuration. So, if we have a d four electronic configuration so that d, four electronic configurations now we can write what we are just telling that it will be t two g, three e g, one and another one will be t two g four only so this will be High spin and the other will be low, spin and terms in terms of the crystal field. Stabilization energy, if we add up all these things related to the corresponding stabilization of the one level as minus four dq or the ah, the corresponding one is two fifth of Δo. This will be minus 3 fifth Δo, but in this particular case the e=-8/5 Δo+t. We have to consider this one pairing energy, because in the t, two g level, we have three things. So, the third electron is going on. All of them are unpaired, but when the fourth electron is coming, it will be paired up in the t2g level, so we should consider this particular p value for this energy difference. So these are the two energy differences for the high spin and the low spin configuration, and we will see what are the corresponding contributory factors which can dictate us that, whether we should have a low spin, complex or a high spin complex, depending upon the nature of the corresponding ligand available to us so p is therefore the corresponding pairing energy and when we consider the pairing energy, together with the crystal field, speeding what we find that it increases as the ligand varies in some order.

Just now what we have seen is that c n minus is greater than f minus and previously, in case of nickel, also, we have seen some order. So, if we put these ligands these three, we have seen this h2o nh3 and en we have seen in case of nickel, 2 plus and in terms of the corresponding equilibrium, the coordination equilibria. We have seen that we can have some order when the nickel is hexaco complex. If we add ammonia, ammonia molecules will be replacing the water molecules, and afterwards, if we put ethylene diamine, that ethylene diamond will also be able to replace the nh3 groups so that we have seen in terms of the corresponding coordination equilibria and that equilibria will tell us That ethylene diamine is a stronger ligand with respect to both water as well as ammonia.

Now we see the magnitudes a little bit. We are now quantifying the magnitude of the Δ, the amount of crystal field speeding, so that energy is basically changing from left to right and we are getting all halides, basically, four halides. If we put then the fluoride is stronger within that particular series of this halide. So, iodide, bromide, chloride and fluoride, so these things are always there and we will find out somewhere that this particular one will also be considering. When we talk about a particular oxidation state of the metal ion. If the metal ion center is oxidizing and if we try to bind with the iodide one, the iodide ligand, and we should also consider the corresponding ability of these centers to remain as the individual form or they are getting oxidized if the metal center is oxidizing. So, there will be a tendency to oxidize iodide ions to iodine chloride ions to chlorine, so that will also lead us to the elimination of iodine bromine or chlorine instead of typical coordination.

But when we see that the fluoride is there and definitely the fluoride is little bit stronger within this particular series and the fluorine will not be there. That means the any metal ion which is forming the corresponding complexes with fluoride will not be able to oxidize the fluoride ion. So, we see that this particular series, so we can put a greater number of uh ligand centers, even in your CBSE book, a huge list of those ligands is there which are commonly encountered by us what we commonly use.

Basically, because already we know this, the halide groups also, these three we have known these are known to us with respect to the binding to the nickel two plus center, and just now we have seen the corresponding efficacy of this cyanide with respect to the fluoride ion. When we talk about their coordination to the trivalent cobalt center, this particular series or any extended series, is known as spectrochemical series so from the left to right. The ligand strength is increasing because that will lead us to some greater amount of splitting.

With respect to the Δ values for these ligands, covering this particular metal ions, so on the left-hand side, the ligands will have the smaller Δ values and the right-hand side will have the bigger large Δ values so the left hand, ligands are weak field ligands and right hand: ligands are strong field ligands. So, just now the way we have seen how we place those electrons in the different d orbitals. So, after d4, we have a situation where we get 85 situations, so d5 situation, the drawing will tell us you see now how we draw this just now. We have seen that the Δ value is less. We get the high spin situation and when the Δ values are high, we have a low spin situation so just change this particular number of electrons. So, this is a high spin situation for a d5 electronic configuration so immediately. We should know also that what d5 electronic configuration we know for the corresponding metal ions, whether it is chromium or manganese or iron in a particular oxidation state, so whether that particular compound will give rise to a corresponding complex, which has a very high magnetic moment or A very low magnetic moment having one unpaired electron only then we have the situation just now what we have seen in case of the cobalt center, so the cobalt, which is trivalent, cobalt and the trivalent cobalt, will therefore definitely be considering in terms of these two situation. Unlike the d5 situation, the d5 situation is that one unpaired electron and five unpaired electrons and in case of d6 it is zero one pair electron and four unpaired electrons. So, four electron Para magnetism versus a diamagnetic situation. So that is the typical drastic condition change that if we are able to make some compound where we get, that means the cobalt we make from a cobalt two salt by adding ammonia and oxidizing it by hydrogen peroxide bar by air, giving rise to corresponding hexamine cobalt. Three complex and that hexamine cobalt three complex is a diamagnetic complex, so just looking at the corresponding magnetic property, we see that the splitting is such that it only prefers the corresponding low spin situation, where all the six electrons will be fulfilling the three t, two g level, giving rise to a corresponding electronic configuration, which will be t two g six then next. The d5 7 situation, which is also true for cobalt, does ion the cobalt two plus ion so cobalt. two+ions will also have a three electron Para magnetism for the wick field. Ligand and for the strong field ligand, we have one electron Para magnetism. So, this is the situation, what we basically encounter for making a compound, which is initially cobalt, two plus like any cobalt. Two salts, like cobalt as chloride or cobalt, as nitrate, if we take in presence of ammonia, is basically getting oxidized by the o2 of air or hydrogen peroxide or any other mild oxidizing engines. We do not need any strong, oxidizing agent to oxidize that and this unpaired electron, which is there if the cobalt center is in the by valence state, will be removed from the system and the system will be oxidized. And this typical stabilization and the diamagnetic compound is getting stabilized in that way. Next is the d8 situation, which is very common for bivalent nickel, so in the nickel two plus situation, what we find that we do not have any condition where we get that thing that in the on the left, also if we even if we go for these Two conditions: we cannot change the number of unpaired electrons on the left, as well as on the right. So, this is a situation where we cannot have any condition that we can put, that is a low spin condition or the high spin condition. So, irrespective of the crystal field will have two unpaired electrons for the bivalent nickel, so these two situations are typically different for d5, d6 and d7, but the di d8 system is completely different. We cannot distinguish between the high spin and the low spin complexes. Then we go to the other field, which is a typical tetrahedral field, so, as we have seen, and we just jotted down all these information’s for the octahedral field now we know how we draw a particular tetrahedral field within a cube, so this particular tetrahedral field when We draw, we have seen that a particular octahedral field or an octahedral crystal field. We have to place six ligands around the metal center, so if, at the center the red dot is the metal ion, we can have the ligands on the six faces of the cube. This is one from the front face and this is another one, the back face, so we have this particular octahedral complex, but what about the tetrahedron one so tetrahedron one? We have to draw again within a cube and we place the same metal ion center at the center of the cube. But now we have the ligands, so ligands will be there. Four ligands will be there at the alternate corners of the cube. So, if we just recall back the shape of the different d orbitals, we now face that in this particular case, the five d orbitals are giving the sets the t two g level and the e g level, but the corresponding interaction for these for the tetrahedral field Will be different where the e g level will be stabilized, and this will be leveled as e and the t2 level will be the destabilized one. So, we will have two low energy levels and three high energy levels and g is dropped, because this has no center of symmetry or center of inversion with respect to the corresponding crystal field. So, this will be the corresponding crystal field splitting when, instead of octahedral field, when we have a corresponding tetrahedral field. So, this we see here for the placement of all the orbitals, so these two will be stabilized, because these are not facing now directly.

These green dots are ligand dots, but in this particular case they interact more with those orbitals. So, this t two set, which is d y j d, z, x and d x y. These three will be destabilized, so t two will be higher in energy compared to the e two e set. So, this is the situation you see now, these the black circles are the alternate corners of the cube and if we consider all the d orbitals available and how they are interacting with each other and more simplistic arrangement, is that this is the metal on center and x y z, if we consider - and these are the approach of the four ligands on the four corners, so this will be the crystal field: speeding with respect to the corresponding spherical environment, when all the four ligands are coming to split these five levels. So, these will be the corresponding levels and likewise our Δ o level. We have the Δt level for the separation is the total separation is Δ t, which will be in the reverse order, so it will be stabilized by three fifth Δt and which will be destabilized by two fifth of Δt, which is reverse of that Of our octahedral geometry, so again we have can have the corresponding electronic configurations from d1to d9, when we place the number of d electrons in these two levels, the number of electrons in the e level and the number of electrons in the t, two Level so we get the corresponding electronic configuration with respect to the crystal field, splitting now the how this magnitude of Δ, and particularly the more examples we consider for the corresponding Δ for the octahedral field, how it is dependent on the other factors.

The first thing we have just considered is the corresponding nature of the corresponding ligand. We have compared that the fluoride, as well as cyanide, so the cyanide ion cn minus, is the stronger ligand compared to fluoride so the electronically. How we consider that thing when we encounter two of the compounds where we can have two different oxidation states of the same metal ion, so the middle ion will also change the corresponding magnitude of this Δ. So, this Δ value, which we experimentally we can determine by measuring the corresponding electronic spectra, because now we have two levels as we. I told you that if we have two levels: e one and e two for electronic transition and if we move the electron from the lower level to the other level, we can experimentally determine the magnitude of that separation between u1 and e2. The same thing we can have with respect to the corresponding hexagon lutheranium, two plus so ruthenium in the by valence state, is giving rise to a nineteen thousand eight-hundred-centimeter inverse separation with respect to the corresponding, ah called, ah wavelength. That means that it can measure it in terms of the corresponding lambda value in nanometer, but if we move it for its corresponding trivalent state, you see is changing from nineteen thousand eight hundred to twenty-eight thousand six-hundred-centimeter inverse so for the same ligand system. That means the same hexa echo species. Our separation is changing with respect to the corresponding oxidation state. So, even if you have this - and if this is not stabilized, this particular species is not stabilized with respect to this ligand, then once it is going for the corresponding oxidation, you can change this oxidation, environment, oxidized, environment by other ligands. As I told you in case of cobalt, cobalt initially has the hexaco cobalt, two plus and which is ultimately oxidized to corresponding cobalt in the trivalent state, which is the hexa amine cobalt. So, the number of ligands and the geometry, as we just see that in octahedral case, we have 6 number of ligands and the corresponding geometry is octahedral and in case of tetrahedron, the number of ligands surrounding the central metal ion is less so the corresponding splitting. That means Δ t will be less than Δ o. That will be there and is roughly if we consider for same type of ligands for the same metal ions in the same oxidation state. We will see that the Δ t so Δ t is about 4 9 of the Δ o values, so these are very weak. So, in most of these cases, when we see that this corresponding fluoride chloride, bromide and iodide species, they are coordinating to the metal ion centers. They are basically giving rise to the corresponding tetrahedral complexes, where there is no such extra stabilization due to the corresponding gain in the cfse values. So, the nature of the ligand, as I told you, that you can expand the corresponding number of ah ligands more, and this has been taken from your CBSE book of chapter of quartz and compounds where we put iodide bromide chloride and fluoride and in between. We are bringing thyroinate as well as sulphide groups also, so the thyroid, when binding to the metal center through nitrogen, will come in between bromide and chloride and the sulphide which is coordinating through sulphur, only which is bigger one and which is little bit soft, also, which Is ah having some strength which is less than fluoride but greater than chloride? Then we occasionally encountered all the oxygen donors: oxygen donors. You see that water is having some greater stabilizing due to the splitting, so the Δ value for water will be higher than that of our hydroxide and though we consider that it has charge similarly for the oxalate ion, it has charge. But the charge is not considered. These are the experimental quantities related to the corresponding Δ values, so the dipoles, the dipole of water molecules, will interact and give the corresponding ligand field, which will be stronger than oxalate ion and the hydroxide ion. And this we already seen that the corresponding one for water, ammonia and ethylenediamine edta will be in between. And lastly, these two things will consider now that why we can place these as the cyanide and the carbonyl complexes that mean the carbon monoxide can also bind to the metal ion in low oxidation state. That will be seen very quickly. So, this gives rise to the corresponding color thing. That means we see that how the crystal field spitting can also explain the color this table is there in your book and how we can read these values. Also, when we have this compound that the pentamine chloro compound of cobalt, two plus so is a bivalent cobalt, so the wavelength, which is absorbed as five thirty-five nanometers, which is in the range of yellow. So, we should remember that the vivjor thing and the vivjor will be the color wheel. We should remember - and we can put the ranges for the corresponding range for the violet indigo blue red etcetera. So, this is the color which is absorbed, but we see the color of the solution, so the complex color will be violet. So, this is the complementary color what we see for the corresponding compound for this complex. Similarly, if we change the chloride ion by water molecule and the oxidation state is also in the trivalent state, our value for this absorption is going to a lower energy value. That means immediately, you can know that one is absorbing at 535 nanometer and the other is absorbing as 500 nanometers.

What does it mean? It is moving to the lower ah values of the wavelength. That means it has high energy value. That means the separation is bigger. Now the Δ value is bigger, so when the Δ value is bigger, it will absorb in the blue green region and the corresponding color of the compound would be red. So definitely this will be the red, and if you go further, that means all of them are substituted by ammonia. No chance of cl so cl is weak. That’s why you can also get the information that cl is weak and force with respect to cl is water? It is stronger and with respect to water, your ammonia is stronger. So now we quantify whatever we have seen earlier, that we can replace those. So, if we get that, if we just have some idea that we can also be able to replace this chloride ion by water and water by ammonia. But it is not so true for compound where we have something where two different types of ligands and pleasure and more complications are there, but simply looking at the molecular formula of these compounds, what we see that the energy is changing towards the high energy values. That means the lower wavelength values. That means your Δ values are changing, so it is going to the range where the absorption of the color is blue and we get the corresponding yellow orange compound, and the solid compound is also nicely orange in color. Yellow is orange in color for this compound, which is nicely crystalline, solid we can get. Similarly, when we compare fluoride as well as this compound, we have compare compared. This is a cyano compound, so hexacyno compound, which is absorbing you see very close to the corresponding. If that is the UV range, so it is 350 is basically the starting point of our visible range, so it is absorbing at 310 nanometers. So, this particular splitting is pretty high and that splitting is basically giving rise to the corresponding range, which is UV range, that we all know. We just know the hydrogen spectrum. We are knowing, in another chapter, the hydrogen spectrum and the separation between the one’s orbital and two s. Orbital of hydrogen is also false in this particular range, which is u v range, but what about the corresponding Δ o values of the octahedral complexes? We see most of them are coming in the visible range, and this visible range is comparable to the series. What we see in case of the hydrogen spectrum is the bomber syringe where we know that the bummer siege, all the electrons’ transitions, are taking place from the higher cells to the second cells. So, the bomber series energy is comparable to the corresponding crystal field energy. For these 3d elements, so this and which is a very paleo, almost colorless in case of the copper case, we have the corresponding. This tetrahedral complex. We can also get the octahedral complexes, which are absorbed at some longer wavelength, so which is red in color. So again, for in respect to copper, these water molecules are giving rise to a wick filled environment, which is red, and the compound is blue and for the hexagon titanium it is 495 nanometer and which is violet in color and which is also characteristic. We can measure these 498 nanometers experimentally that we just see so a light of wavelength. We have to measure with respect to the transition from the e1 to e2 and that transition, if we just apply how this transition can take place with respect to corresponding hexa echo titanium compound, which is trivalent. So just now. What we have seen from your book is telling us that it will have some absorption as 598 nanometers, so a range of light, which is in the 500 nanometers, which is easy to remember that the 500-nanometer energy is sufficient to promote the electron from one level. To the other, which is the ground level to the excited level for one electron, which is in the t2g level, to the eg level. So, this transition can take place due to the light absorption of 500 nanometer, and we can also measure the corresponding Δ value. For that transition so how, it looks like we basically measure in one axis, it is the corresponding centimeter inverse. We have plotted, but experimentally, in a spectrometer, we ah measure in the nanometer scale. So, this is the wavelength axis, and this is the absorbance axis, so that basically gives rise to the corresponding maximum absorption with respect to 498 nanometers, and that 498 nanometer is equivalent to 20 300-centimeter inverse is due to the transition from the corresponding level, which we See that the transition is at excite, so it is upper level. Is the eg so easy to t2g energy gap, or you can consider it as the transition when you move when you promote this electron to the other, so the transition is when we, after sometime it will come down. So, you can write this transition as also and during absorption. What happens is the absorption spectroscopy optical absorption. We are talking so the during absorption. The transition is taking from t to g to a g, but after some time it relaxes from eg to t2g an energy gap in terms of the wavelength which is 498 nanometer in terms of centimeter inverse and in terms of kilojoule per mole. Also, we can have - which is 243 kilojoule per mole, so this particular information that when we have a metal ion in solution, we see the color and we see the transitions very nicely. So, what about the thing which we see for the color of some gemstones, which is also there a part, a one page? Basically, in your book - and you should read it nicely there - that when a particular light of visible range can ah strike a particular material which is the precious gemstone, it will absorb some color like your solution, the solution is absorbing one part of the color and its Eliminating or giving rise to the corresponding color, which is complementary in nature. So, what is ruby? So, ruby has a very fine color, which is red in color, and it absorbs all other wavelengths from the white light spectrum only it is reflecting the red. That means it is the complementary color which is coming out from that is red. So that is why ruby is red in color and ruby is nothing but is a gem quality. Corundum quantum is nothing but our crystalline alumina al2o3, but the color is due to some impurity, which we call as the doping. So, one percent or less than one percent point five to one percent. Doping of the chromium three plus on the corundum can give rise to the chromium three plus electronic spectra for the corresponding color. What is happening now in the solid state? All the oxides are your new ligands, so in the solid state we can consider o2 minus. As our new ligands instead of water or hydroxide ion, so these o2 minus will be now placing around the chromium, 3 plus and will just distort the octahedral arrangement, which was originally present for the alumina structure, because the chromium size is different from the aluminum size. So doping is basically bringing some information to the system and it will slightly alter the corresponding absorption, which we do not get for the corresponding hexa ako chromium 3 complexes. So, the new position of the o2 minus in the distorted form will be responsible for a typical coloration for this ruby gem. Another one is also sapphire and is the chemical composition of both of them are corundum, but they exhibit different colors. This is due to the placement of the corresponding one, which we get, that the composition is different, so primary chemical composition may be same, but the corresponding impurities are different. So, what is purple in color? The sapphire one is purple where we get instead of chromium in ruby, we have vanadium and it can come from different shades depending upon the corresponding distortion and sometime. You can have some iron also present in this particular thing and you can have the paleolithic green in color. So, if both titanium and iron impurities are present together and a correct valence state. But what is the oxidation state of the titanium center and what is the corresponding oxidation state of the iron center? We will be able to get a deep, blue, color so synthetically, also in the laboratory. Nowadays we can make synthetic gem materials or gemstones by knowing the corresponding metal ion, which we can impregnate and we get the corresponding coloration for all these things. So, beryl is another example for emerald is a colorless pure mineral if it is pure only, but when chromium impurity like ruby is there, it will have a different color, so its manganese is added. Instead of chromium, burial becomes pink and name will be morganite, but if iron is present, this will be a different color and becomes an aquamarine, so it is blue is in color. So, all these things are basically related to the formation of this beryllium in the aluminum cyclosilicate. So, what we talk about now, like your corundum here also we have aluminum at the side, but is not in oxide lattice, but is a silicate lattice cycle, cyclic silicate lattice. We have, and we try to replace this aluminum by chromium, this aluminum by manganese and this aluminum by iron, because all are 3d elements, so 3d elements can be useful to substitute this aluminum, which is in the octahedral field, having comparable size. So, the supplier that replaced this mrn will give you a green coloration for this. When we have trace amount of chromium and sometime, we can have vanadium also. So, this impurity on all these gems will give rise to something which can give rise to the corresponding compounds. So, like your valence bond theory, the crystal field theory also has some limitations, but it can explain many more things which we cannot explain in the case of the corresponding valence bond theory. So here we take the very basic assumption that we are considering these. As point charges, but all these ligands, whatever ligands we can have, that cannot be the point charges. So, whatever is this? We are talking here in this environment. These ligands, which we are talking about as point charges, but is not that if we have a bigger ligand say if you have an iodide is also not a point charge and if you have a very bigger organic molecule or organic moiety over there. So, this will not be a point charge, so this particular problem comes to us when we talk about the corresponding ligand as carbon monoxide, carbon monoxide is a well-known ligand and long back. It was discovered that we can make some compounds during the purification of nickel, the solid nickel which is in the atomic state. That means the nickel zero when the gas valve contains nickel and the gas cylinder contains carbon monoxide, it is getting corroded due to the formation of tetracarbonyl nickel zero compound. So, this is the corresponding deposition on the gas cylinder with that valve, and that valve will tell us this can form and then in the zero-oxidation state. So, this is one aspect that how we can tackle a corresponding compound where the nickel is in zero oxidation state. Another one is the corresponding ligand, which is not a point charge. So, if we do not consider that point charge, so the typical interaction, what we are discussing so far about the corresponding charges for the positively charged metal ion and the negatively charged ligand that the interaction we are considering as the corresponding interaction is purely electrostatic. That means the electrostatic interaction, what we see in the rock salt like sodium chloride, but that electrostatic interaction is not present.

What we see here is that, if we have some molecule like carbon monoxide or c n minus another example, is also c n minus. So, it will have some orbitals, so those orbitals will have some ah lone pair of electrons, and this metal ion will also have some orbitals, which are having some vacancy or field. So, there will be something where we can overlap the orbitals from the metal ion and the orbitals of the ligand. So, this particular picture of the electrostatic interaction only picture is now slowly fading away. So, we have to consider some amount of covalent interaction so that amount of covalent interaction how we can modify in case of a typical complex, which is like m l six. So, what is that covalent interaction? Covalent interaction is nothing but molecular orbital formation. What we have seen like that of the same formation of carbon monoxide molecule how a carbon monoxide molecule can form out of that corresponding configuration like this Lewis, dot structure for the carbon monoxide molecule from the atomic orbitals of carbon and the from the atomic orbitals of Oxygen, so what we get at the end, we get the corresponding molecular orbitals of m of carbon monoxide now the donor levels or the acceptor levels, whatever we have on the ligand. Now it is your ligand, so this ligand will have a certain number of molecular orbitals and those molecular orbitals will now slowly interact with the atomic orbitals of this metal ion. But when we have, this is also a picture. What we will get like the formation of carbon monoxide if m, on the left, hand, side and l on the right-hand side. So, we can also draw ah some levels, which will be your molecular orbital levels like this. One will be stabilized and there will be a destabilized molecular orbital level for the m l six, so that is the more improved version of the crystal field theory for bonding picture. So, we will not be able to consider this as the corresponding point charges for the dipole. Only case and which does not take into account of the overlap of the ligand and the metal orbitals so consequence, is that why we can say that the carbon monoxide is the stronger ligand, then cyanide, we will unable to explain that antenna unless we consider that carbon Monoxide is having some interaction with the metal center as in covalent nature, so we go for a ligand field. Theory where we introduce the concept of molecular orbital picture for this and the valence orbitals will be considered by involving the middle ion orbitals and the ligand orbitals are symmetry adapted. Linear combination, salc is nothing, but you remain within the symmetry picture that symmetry adapted linear combinations of those orbital will give rise to the several molecular orbitals, and those molecular orbitals will give rise to types of bonding which can be sigma bonding and which can also be Pi bonding, so we can, where we have the ligand as a single valence orbital directed towards the central of the metal ion and the pi bonding when the ligand has the field orbital of the pi symmetry around the metal ligand axis, whether middle is, can give you the function as a donor or the ligand can also function as a donor. So, this we see for a very good example that the corresponding molecular orbital detail molecular orbital picture, which is there in any common book that what we see that these are the levels and instead of writing this, that is the carbon which is having higher energy. For this two s and two p level compared to this oxygen. So, when we have the total number of ten electrons on carbon monoxide, we will be having the placement of this as the three-sigma electron and two pi electrons. So, the sigma electrons on have some character, which is close to the carbon character, which will give rise to the corresponding sigma donation. So, the homo will have this particular picture and this blue orbital, which is on the carbon side. This is carbon, and this is oxygen, so this will be the donor orbital. Similarly, the lumo. Basically, we can have two lumo. Two of them are of the pi character, so these two-pi character of this, so the lumo of this will be there. So, on the ah carbon monoxide side we can have this loma also will be there and available for bonding to the metal ion center or the corresponding metal in the zero-oxidation state. So, this is typically the sigma donation and that sigma donation will basically give rise to something where the carbon side of this will be bound to the corresponding metal ion. Just now what we have seen in case of nickel in the zero-oxidation state, interacting with the carbon monoxide, so we will have four such bonds will be forming. So, the nickel carbon bond will be there, so we can have four nickel carbon bonds in tetra carbonyl, nickel, zero species, and if we consider this nickel is in the zero-oxidation state, we can have all the levels filled. So, you cannot have a 3d 8 situation. We can have a 3d 10 situation, so all the levels are filled still some of the orbitals which are higher in energy or some of the molecular orbitals. Really speaking, some of the molecular orbitals will all be available which are high energy, acceptor orbitals, so high energy acceptor orbitals will be there which can accept the electron density. The sigma donation, if we consider from the carbon monoxide to those level for a typical one direction bonding, but at the same time, since the all the levels are filled on the nickel zero nickel zero. If you have the orbitals, whether it is a typical, ah atomic orbital, but we are not considering atomic orbitals now these are the molecular orbitals on metal, centered or molecular orbital, so the field orbital electrons. The orbitals now push the electron density to the empty molecular orbitals on the carbon monoxide, so highest unoccupied, molecular orbitals, so the highest unoccupied molecular orbital. The homos are available for accepting the electron density from the corresponding metal centers. So, this will give you one corresponding sigma donation, and this will be considered as pi acceptance. So that is why the carbon monoxide molecules will be considered as good pi acceptor ligands. We classify them pi, acceptor ligands, and that pi acceptor ligands will give rise to some multiple bonding between the metal and the carbon center, and the interaction is pretty strong and that strong interaction is basically responsible for changing the corresponding splitting between the levels and with the Splitting in the terms of the corresponding Δ values, what we are considering for the crystal field theory, so the separation is very high so on the spectrochemical series that’s why the carbon monoxide is, on the extreme right hand, side. So, in the zero-oxidation state we can have several such compounds, and these compounds are very good examples for the simple carbon monoxide as the ligand for these compounds. So, we have the nickel and this nickel is coordinating to those co which is tetrahedral in nature, and this tetrahedral cos are there and this basically another co. So, these four co will be giving rise to this now, the stabilization, so the stabilization we can consider little bit with respect to the effective atomic number. What 18-electron rule for the main group compounds with respect to the 18-electron rule, so the 18 electron rules can also be applied to these species such that initially, we should know what the formula is. So, this is the carbonyl compound. Unlike your n, i c n whole four, two minus where this nickel is plus two. So, if we count the total number of electrons, this will be eight, and this is giving rise to four into two. So, four plus is a sixteen electron species, so it is not eighteen electron species so, but it has some stabilization in a particular geometry, because we all know that in a particular geometry is for this is the square, planar geometry, so the square planar geometry in this Particular environment: it has the stabilization, but for the zero-oxidation state, this nickel will have now 10 electrons plus the cobalt. Carbon monoxide also provides two electrons, so four into two, so is eighteen electron system, so tetracarbon in nickel is an eighteen-electron system. So, it has a stabilization if we consider not ten instead of ten. If we consider that another eighteen electrons from the whole series will be there for the effective atomic number, for this will be 16 in that particular case. Also, not only nickel tetracarbonyl, you can also consider it for the iron iron has some electron, which is there. We know that the eight which is atomic number of twenty-six, so twenty-six. That means eight plus two into five ten. So, it is also eighteen electrons. This is also eighteen electrons, but the situation is that what you can have for this dimeric compound this. These are also these two dimeric examples which is the corresponding manganese dimer and the cobalt dimer. What is there in your book? Also, this is the last slide. What I have taken from your book also - and you should also have some idea about the stabilization, so these are the five carbonyl compounds. These are the good example of your corresponding organometallic compound, though they have given as the corresponding carbonyl compound where the carbonyl ligand is, in the extreme right hand, side of the spectrochemical series, so this is 18 electron configurations. This has 18 electrons, and this has also 18 electron configuration this chromium, because the chromium having the 6 electron and the 6-carbon monoxide can give rise to six into two twelve electrons. So, this is also another eighteen electron species. But what about this manganese? Basically, manganese when we see that manganese, we get so manganese is zero, so that will give rise to a 7 electron to you and then around 5. That means we are not going for an octahedral situation, so we get 5 for that. So that is there. So, we get in this fashion, so 5 of them, so five co, so five into two so is equal to ten. So, all together we are getting a 17 electron species, so this 17 electron species is not stable. So, if it can get something where we can have some bonding with another fragment, so this is one part so left-hand Part of this. If we go for another part like this m n c, o whole five. So, due to this manganese bonding, we have to consider one electron because it is made up of two electrons, so this manganese bonding can contribute another electron. So, this will also be eighteen electrons, so that basically gives that to the corresponding stable species, which is eighteen electrons in nature. Similarly, for the species, like c o, two c o hole eight, so that will also give rise to the corresponding counting and all the time. We should have some idea that how much carbon monoxide is there, five will be as mono rented and no such as the bridging groups and one as the metal bonding. Similarly, for the corresponding cobalt system, you can have the bridging carbon monoxide, because this can also function as a very good bridging group, but considering the number of electrons for this particular one. We can also consider the bond, so this will also have a cobalt bond, which will consider that extra electron giving rise to a corresponding 18 electron configuration for that species. So, all these compounds, whether we have the three mononuclear compound nickel, iron and chromium, but two di nuclear compounds the manganese or the cobalt, because in this particular case we have the cobalt bond.

Further You have the corresponding bridging, because the number of cobalt centers is less compared to your manganese compound. So, this is bridging, because the bo all the center, because the left-hand cobalt center and the right hand, cobalt centers, are octahedral in nature. So, all these five electrons basically are stable and we can have some idea in relation to the 18-electron rule. So, if is something is not known, so f is unknown and your ligand also tells us that it is carbon monoxide.

We all know that these can be achieved if we have two energy levels, one is e1 and another is e2 and there is an electronic transition from one particular level to another level due to the absorption of h, and that h, will have some relationship with the corresponding lambda value, so the one lambda will be absorbed, so we will have the absorbed lambda and we see the corresponding complementary color.

So how many co? We can accommodate around the metal center is easy to decipher easy to identify from applying this 18-electron rule, whether it is nickel or iron or chromium or a dimeric species. By looking at the 18-electron configuration with respect to all these species and that stability is basically gained out of that, and we can have some idea, that is not a simple example and you have to memorize everything logically. You have to think that 18 electron configuration is preserved in all these species, and the available number of carbon monoxide is there. So definitely if m is covered. Your carbon monoxide number is eight. Thank you very much.